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Next: Exercises Up: Equilibrium of Compressible Fluids Previous: Atmospheric Stability

Eddington Solar Model

Let us investigate the internal structure of the Sun, which is basically a self-gravitating sphere of incandescent ionized gas (consisting mostly of hydrogen). Adopting a spherical coordinate system (see Section C.4), $ r$ , $ \theta $ , $ \phi$ , whose origin coincides with the Sun's geometric center, and making the simplifying (and highly accurate) assumption that the mass distribution within the Sun is spherically symmetric, we find that

$\displaystyle \frac{dm}{dr}= 4\pi\,r^{\,2}\,\rho,$ (13.21)

where $ m(r)$ is the total mass contained within a sphere of radius $ r$ , centered on the origin, and $ \rho(r)$ the mass density at radius $ r$ . Now, as is well known, the gravitational acceleration at some radius $ r$ in a spherically symmetric mass distribution is the same as would be obtained were all the mass located within this radius concentrated at the center, and the remainder of the mass neglected (Fitzpatrick 2012). In other words,

$\displaystyle \frac{d{\mit\Psi}}{dr} = \frac{G\,m}{r^{\,2}},$ (13.22)

where $ {\mit\Psi}(r)$ is the gravitational potential energy per unit mass, and $ -d{\mit\Psi}/dr$ the radial gravitational acceleration. The force balance criterion (13.1) yields

$\displaystyle \frac{dp}{dr} + \rho\,\frac{d{\mit\Psi}}{dr} = 0,$ (13.23)

where $ p(r)$ is the pressure. The previous three equations can be combined to give

$\displaystyle \frac{1}{r^{\,2}}\frac{d}{dr}\!\left(\frac{r^{\,2}}{\rho}\,\frac{dp}{dr}\right) = -4\pi\,G\,\rho.$ (13.24)

In order to make any further progress, we need to determine the relationship between the Sun's internal pressure and density. Unfortunately, this relationship is ultimately controlled by energy transport, which is a very complicated process in a star. In fact, a star's energy is ultimately derived from nuclear reactions occurring deep within its core, the details of which are extremely complicated. This energy is then transported from the core to the outer boundary via a combination of convection and radiation. (Conduction plays a much less important role in this process.) Unfortunately, an exact calculation of radiative transport requires an understanding of the opacity of stellar material, which is an exceptionally difficult subject. Finally, once the energy reaches the boundary of the star, it is radiated away. The following ingenious model, due to Eddington (Eddington 1926), is appropriate to a star whose internal energy transport is dominated by radiation. This turns out to be a fairly good approximation for the Sun. The main advantage of Eddington's model is that it does not require us to know anything about stellar nuclear reactions or opacity.

Now, the temperature inside the Sun is sufficiently high that radiation pressure cannot be completely neglected with respect to conventional gas pressure. In other words, we must write the solar equation of state in the form

$\displaystyle p = p_g + p_r,$ (13.25)


$\displaystyle p_g = \frac{\rho\,k\,T}{\mu\,m_p}$ (13.26)

is the gas pressure (modeling the plasma within the Sun as an ideal gas of free electrons and ions), and (Chandrasekhar 1967)

$\displaystyle p_r = \frac{1}{3}\,\alpha\,T^{\,4}$ (13.27)

the radiation pressure (assuming that the radiation within the Sun is everywhere in local thermodynamic equilibrium with the plasma). Here, $ T(r)$ is the Sun's internal temperature, $ k$ the Boltzmann constant, $ m_p$ the mass of a proton, and $ \mu$ the relative molecular mass (i.e., the ratio of the mean mass of the free particles making up the solar plasma to that of a proton). Note that the electron mass has been neglected with respect to that of a proton. Furthermore, $ \alpha=4\,\sigma/c$ , where $ \sigma$ is the Stefan-Boltzmann constant, and $ c$ the velocity of light in a vacuum. Incidentally, in writing Equation (13.26), we have expressed $ M/R$ in the equivalent form $ \mu\,m_p/k$ .


$\displaystyle p_g$ $\displaystyle =(1-\beta)\,p,$ (13.28)
$\displaystyle p_r$ $\displaystyle = \beta\,p,$ (13.29)

where the parameter $ \beta $ is assumed to be uniform. In other words, the ratio of the radiation pressure to the gas pressure is assumed to be the same everywhere inside the Sun. This fairly drastic assumption turns out--perhaps, somewhat fortuitously (Mestel 1999)--to lead to approximately the correct internal pressure-density relation for the Sun. In fact, Equations (13.26)-(13.29) can be combined to give

$\displaystyle p = K\,\rho^{\,4/3},$ (13.30)


$\displaystyle K = \left[\left(\frac{k}{m_p}\right)^4\,\frac{3}{\alpha}\,\frac{\beta}{\mu^{\,4}\,(1-\beta)^4}\right]^{1/3}.$ (13.31)

It can be seen, by comparison with Equation (13.4), that the previous pressure-density relation takes the form of an adiabatic gas law with an effective ratio of specific heats $ \gamma=4/3$ . Note, however, that the actual ratio of specific heats for a fully ionized hydrogen plasma, in the absence of radiation, is $ \gamma=5/3$ . Hence, the $ 4/3$ exponent, appearing in Equation (13.30), is entirely due to the non-negligible radiation pressure within the Sun.

Figure: The functions $ \theta (\xi )$ (solid) and $ y(\xi )$ (dashed).

Let $ T_c = T(0)$ , $ \rho_c= \rho(0)$ , and $ p_c= p(0)$ , be the Sun's central temperature, density, and pressure, respectively. It follows from Equation (13.30) that

$\displaystyle p_c = K\,\rho_c^{4/3},$ (13.32)

and from Equations (13.26) and (13.28) that

$\displaystyle T_c = \frac{p_c}{\rho_c}\,\frac{m_p}{k}\,(1-\beta)\,\mu.$ (13.33)

Suppose that

$\displaystyle \frac{T}{T_c}= \theta,$ (13.34)

where $ \theta(r)$ is a dimensionless function. According to Equations (13.26), (13.28), and (13.30),

$\displaystyle \frac{\rho}{\rho_c}$ $\displaystyle = \theta^{\,3},$ (13.35)
$\displaystyle \frac{p}{p_c}$ $\displaystyle = \theta^{\,4}.$ (13.36)

Moreover, it is clear, from the previous expressions, that $ \theta=1$ at the center of the Sun, $ r=0$ , and $ \theta=0$ at the edge, $ r=R$ (say), where the temperature, density, and pressure are all assumed to fall to zero. Suppose, finally, that

$\displaystyle r = a\,\xi,$ (13.37)

where $ \xi$ is a dimensionless radial coordinate, and

$\displaystyle a = \left(\frac{K}{\pi\,G\,\rho_c^{\,2/3}}\right)^{1/2}.$ (13.38)

Thus, the center of the Sun corresponds to $ \xi=0$ , and the edge to $ \xi=\xi_1$ (say), where $ \theta(\xi_1)=0$ , and

$\displaystyle R = \xi_1\,a.$ (13.39)

Figure 13.2: Solar temperature versus mass fraction obtained from the Eddington Solar Model (solid) and the Standard Solar Model (dashed).

Equations (13.35)-(13.38) can be used to transform the equilibrium relation (13.24) into the non-dimensional form

$\displaystyle \frac{1}{\xi^{\,2}}\frac{d}{d\xi}\!\left(\xi^{\,2}\,\frac{d\theta}{d\xi}\right) = -\theta^{\,3}.$ (13.40)

Moreover, Equation (13.21) can be integrated, with the aid of Equations (13.35), (13.37), and (13.40), and the physical boundary condition $ m(0)=0$ , to give

$\displaystyle m = 4\pi\,\rho_c\,a^{\,3}\,y,$ (13.41)


$\displaystyle y(\xi) = -\xi^{\,2}\,\frac{d\theta}{d\xi}.$ (13.42)

Equation (13.40) is known as the Lane-Emden equation (of degree $ 3$ ), and can, unfortunately, only be solved numerically (Chandrasekhar 1967). The appropriate solution takes the form $ \theta= 1-\xi^{\,2}/6 + {\cal O}(\xi^{\,4})$ when $ \xi\ll 1$ , and must be integrated to $ \xi=\xi_1$ , where $ \theta(\xi_1)=0$ . Figure 13.1 shows $ \theta (\xi )$ , and the related function $ y(\xi )$ , obtained via numerical methods. Note that $ \xi_1=6.897$ , and $ y_1=y(\xi_1) = 2.018$ .

Figure 13.3: Solar mass density versus mass fraction obtained from the Eddington Solar Model (solid) and the Standard Solar Model (dashed).

According to Equation (13.41), the solar mass, $ M_\ast= m(R)$ , can be written

$\displaystyle M_\ast = 4\pi\,\rho_c\,a^{\,3}\,y_1,$ (13.43)

which reduces, with the aid of Equations (13.31) and (13.38), to

$\displaystyle \frac{\beta}{(1-\beta)^4} = \zeta^{\,2},$ (13.44)


$\displaystyle \zeta = \mu^{\,2}\,\frac{M_\ast}{M_0},$ (13.45)


$\displaystyle M_0 = \left[\frac{1}{(\pi\,G)^3}\left(\frac{k}{m_p}\right)^4\,\frac{3}{\alpha}\right]^{1/2}4\pi\,y_1 = 3.586\times 10^{31}\,{\rm kg}.$ (13.46)

Moreover, it is easily demonstrated that

$\displaystyle K = \frac{\pi\,G\,M_\ast^{\,2/3}}{(4\pi\,y_1)^{2/3}}.$ (13.47)

According to Equations (13.44) and (13.45), the ratio, $ \beta/(1-\beta)$ , of the radiation pressure to the gas pressure in a radiative star is a strongly increasing function of the stellar mass, $ M_\ast$ , and mean molecular weight, $ \mu$ . In the case of the Sun, for which $ \zeta\ll 1$ , Equation (13.44) can be inverted to give the approximate solution

$\displaystyle \beta = \zeta^{\,2} - \frac{1}{4}\,\zeta^{\,4} + {\cal O}(\zeta^{\,6}).$ (13.48)

Using the observed solar mass $ M_\ast=1.989\times 10^{30}\,{\rm kg}$ , and the value $ \mu=0.68$ (which represents the best fit to the Standard Solar Model mentioned in the following), we find that $ \beta=6.58\times 10^{-4}$ . In other words, the radiation pressure inside the Sun is only a very small fraction of the gas pressure. This immediately implies that radiative energy transport is much less efficient than convective energy transport. Indeed, in regions of the Sun in which convection occurs, we would expect the convective transport to overwhelm the radiative transport, and so to drive the local pressure-density relation toward an adiabatic law with an exponent $ 5/3$ . Fortunately, convection only takes place in the Sun's outer regions, which contain a minuscule fraction of its mass.

Figure 13.4: Solar pressure versus mass fraction obtained from the Eddington Solar Model (solid) and the Standard Solar Model (dashed).
\epsfysize =3in

Equations (13.32), (13.33), (13.39), (13.43), and (13.47) yield

$\displaystyle T_c$ $\displaystyle = \frac{\xi_1}{4\,y_1}\,\mu\,(1-\beta)\,\frac{G\,M_\ast\,m_p}{R\,k} = 1.34\times 10^7\,{\rm K},$ (13.49)
$\displaystyle \rho_c$ $\displaystyle = \frac{\xi_1^{\,3}}{4\pi\,y_1}\,\frac{M_\ast}{R^{\,3}} = 7.63\times 10^4\,{\rm kg\,m^{-3}},$ (13.50)
$\displaystyle p_c$ $\displaystyle = \frac{\xi_1^{\,4}}{16\pi\,y_1^{\,2}}\,\frac{G\,M_\ast^{\,2}}{R^{\,4}}= 1.24\times 10^{16}\,{\rm N\,m^{-2}},$ (13.51)

where the solar radius $ R$ has been given the observed value $ 6.960\times 10^8\,{\rm m}$ . The actual values of the Sun's central temperature, density, and pressure, as determined by the so-called Standard Solar Model (SSM),13.1which incorporates detailed treatments of nuclear reactions and opacity, are $ T_c = 1.58\times 10^7\,{\rm K}$ , $ \rho_c = 15.6\times 10^4\,{\rm kg\,m^{-3}}$ , and $ p_c= 2.38\times 10^{16}\,{\rm
N\,m^{-2}}$ , respectively. It can be seen that the values of $ T_c$ , $ \rho_c$ , and $ p_c$ obtained from the Eddington model lie within a factor of two of those obtained from the much more accurate SSM. Figures 13.2, 13.3, and 13.4 show the temperature, density, and pressure profiles, respectively, obtained from the SSM13.2 and the Eddington model. The profiles are plotted as functions of the mass fraction, $ m(r)/M_\ast = y(\xi)/y_1$ , where $ \xi=r/a$ . It can be seen that there is fairly good agreement between the profiles calculated by the two models. Finally, Figure 13.5 compares the ratio, $ \beta/(1-\beta)$ , of the radiation pressure to the gas pressure obtained from the SSM and the Eddington model. Recall, that it is a fundamental assumption of the Eddington model that this pressure ratio is uniform throughout the Sun. In fact, it can be seen that the pressure ratio calculated by the SSM is not spatially uniform. On the other hand, the spatial variation of the ratio is fairly weak, except close to the edge of the Sun, where convection sets in, and the Eddington model, thus, becomes invalid. We conclude that, despite its simplicity, the Eddington solar model does a remarkably good job of accounting for the Sun's internal structure.

Figure 13.5: Ratio of radiation pressure to gas pressure calculated from the Eddington Solar Model (solid) and the Standard Solar Model (dashed).

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Next: Exercises Up: Equilibrium of Compressible Fluids Previous: Atmospheric Stability
Richard Fitzpatrick 2016-03-31